HDU 1240 (简单三维广搜) Asteroids!
给出一个三维的迷宫以及起点和终点，求能否到大终点，若果能输出最短步数
三维的问题无非就是变成了6个搜索方向
最后强调一下xyz的顺序，从输入数据来看，读入的顺序是map[z][x][y]
总之，这是很基础的一道题
1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <queue> 6 #include <algorithm> 7 using namespace std; 8 9 struct Point 10 { 11 char type; 12 int x, y, z; 13 int steps; 14 }start, end; 15 16 char map[12][12][12]; 17 int dir[6][3] = {{1,0,0}, {-1,0,0}, {0,1,0}, {0,-1,0}, {0,0,1}, {0,0,-1}}; 18 char buff[10]; 19 int n; 20 21 bool islegal(int x, int y, int z) 22 { 23 return (x>=0 && x<n && y>=0 && y<n && z>=0 && z<n && map[z][x][y]!='X'); 24 } 25 26 void BFS(void) 27 { 28 queue<Point> qu; 29 start.steps = 0; 30 qu.push(start); 31 while(!qu.empty()) 32 { 33 Point now = qu.front(); 34 if(now.x==end.x && now.y==end.y && now.z==end.z) 35 { 36 printf("%d %d\n", n, now.steps); 37 return; 38 } 39 for(int i = 0; i < 6; ++i) 40 { 41 int xx = now.x + dir[i][0]; 42 int yy = now.y + dir[i][1]; 43 int zz = now.z + dir[i][2]; 44 if(islegal(xx, yy, zz)) 45 { 46 Point next; 47 next.x = xx, next.y = yy, next.z = zz, next.steps = now.steps + 1; 48 map[zz][xx][yy] = 'X'; 49 qu.push(next); 50 } 51 } 52 qu.pop(); 53 } 54 printf("NO ROUTE\n"); 55 } 56 57 int main(void) 58 { 59 #ifdef LOCAL 60 freopen("1240in.txt", "r", stdin); 61 #endif 62 63 while(cin >> buff >> n) 64 { 65 for(int i = 0; i < n; ++i) 66 for(int j = 0; j < n; ++j) 67 cin >> map[i][j]; 68 69 cin >> start.x >> start.y >> start.z; 70 cin >> end.x >> end.y >> end.z; 71 cin >> buff; 72 BFS(); 73 } 74 return 0; 75 }
